How do you find the derivative of #g(t) = 1/t^(1/2)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Alan N. Mar 1, 2017 #g'(t) = -1/(2t^(3/2))# Explanation: #g(t) = 1/t^(1/2) = t^(-1/2)# Applying the power rule: #g'(t) = -1/2 t^((-1/2-1)) = -1/2 t^(-3/2)# #= -1/(2t^(3/2))# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1092 views around the world You can reuse this answer Creative Commons License